Chhetri AcademyGCSE & A level Paper Builder

N5Listing and the product rule

Edexcel GCSE Maths (1MA1), Higher tier · Number

Practise Listing and the product rule. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

Build a paper on this topic

▶ Watch videos on Listing and the product rule (Corbettmaths on YouTube) · Practise all of Number

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
Spinner A is numbered 1, 2 and 3. Spinner B is numbered 1, 2 and 3.
Sam spins both spinners and adds the two numbers to get a total.
(a) List all the possible outcomes, writing the number on spinner A first.[2]
(b) How many of the outcomes give a total of 4?[1]
Show the answer and mark scheme
(a) Answer: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)
  • B2 for all 9 outcomes and no extras

Worked solution: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)

(b) Answer: 3
  • B1 for 3 (ft their list)

Worked solution: (1, 3), (2, 2), (3, 1)

Question 2Medium3 marks
A password is made of 3 letters from A to Z followed by 2 digits from 0 to 9.
The letters can be repeated. The two digits must be different.
Ravi says, 'The number of possible passwords is \(26^3 \times 10^2\).'
(a) Explain the mistake in Ravi's method.[1]
(b) Work out the number of possible passwords.[2]
Show the answer and mark scheme
(a) Answer: \(10^2\) allows the two digits to be the same; once the first digit is chosen there are only 9 choices for the second.
  • C1 for explaining that there are only 9 choices for the second digit (\(10^2\) includes repeated digits such as 33)

Worked solution: The two digits must be different, so the second digit has 9 choices, not 10.

(b) Answer: 1 581 840
  • M1 for \(26^3 \times 10 \times 9\) oe
  • A1 for 1 581 840

Worked solution: \(26 \times 26 \times 26 \times 10 \times 9 = 17\,576 \times 90 = 1\,581\,840\)

Question 3Hard4 marks
How many whole numbers from 1000 to 9999 are even and have four different digits?
You must show your working.[4]
Show the answer and mark scheme
Answer: 2296
  • P1 for splitting into cases: last digit 0, and last digit 2, 4, 6 or 8
  • P1 for \(9 \times 8 \times 7\) (= 504) when the last digit is 0
  • P1 for \(8 \times 8 \times 7\) (= 448) for each of the other even last digits, or \(4 \times 448\) (= 1792)
  • A1 for 2296

Worked solution: Last digit 0: the first digit has 9 choices (1–9), then 8 and 7 choices: \(9 \times 8 \times 7 = 504\).
Last digit 2, 4, 6 or 8: the first digit cannot be 0 or the last digit, so 8 choices, then 8 and 7: \(8 \times 8 \times 7 = 448\) each, \(4 \times 448 = 1792\).
Total \(504 + 1792 = 2296\).

Related subtopics

Stuck? Get 1-to-1 help. Chhetri Academy tutors GCSE and A level Maths and Science online, with a free 30-minute trial lesson.

Book a free trial